A car was seen driving off the edge of a cliff at an angle of 10.° above the horizontal.
The cliff is 95. m high and the car impacted the ground at a distance of 156. m from the base of the cliff.
The speed of the car as it left the cliff was closest to:
To find the speed of the car as it left the cliff, we can use the kinematic equations of motion.
First, let's break down the given information:
- Height of the cliff (h) = 95 m
- Horizontal distance from the base of the cliff (d) = 156 m
- Angle of the car's trajectory (θ) = 10°
Now, we can use the following equations of motion to solve for the initial velocity of the car (v₀):
1. Vertical displacement equation:
h = v₀sin(θ)t - (1/2)gt²
2. Horizontal displacement equation:
d = v₀cos(θ)t
3. Time equation (from the horizontal displacement equation):
t = d / (v₀cos(θ))
Substituting the time equation into the vertical displacement equation, we get:
h = (v₀sin(θ))(d / (v₀cos(θ))) - (1/2)g(d² / (v₀²cos²(θ)))
h = dsin(θ) / cos(θ) - (1/2)g(d² / v₀²cos²(θ))
hcos(θ) = d - (1/2)g(d² / v₀²cos²(θ))
hcos³(θ) = dcos(θ) - (1/2)gd² / v₀²cos²(θ)
hcos³(θ) = dcos(θ) - (1/2)gd² / (v₀²cos²(θ))
h²cos⁴(θ) = d²cos²(θ) - (1/2)gd² / v₀²
2h²cos⁴(θ) = 2d²cos²(θ) - gd² / v₀²
2h²cos⁴(θ) - 2d²cos²(θ) = gd² / v₀²
v₀² = gd² / (2h²cos⁴(θ) - 2d²cos²(θ))
Finally, we can substitute the given values to find the speed of the car as it left the cliff:
g = 9.8 m/s² (acceleration due to gravity)
h = 95 m
d = 156 m
θ = 10°
v₀² = (9.8 m/s²)(156 m)² / (2(95 m)²cos⁴(10°) - 2(156 m)²cos²(10°))
To find the speed of the car as it left the cliff, we need to use kinematic equations and apply basic trigonometry.
Given:
Angle above the horizontal (θ) = 10.°
Height of the cliff (h) = 95. m
Horizontal distance traveled (x) = 156. m
First, let's break down the initial velocity of the car into its horizontal and vertical components. The horizontal component remains constant (v_x) while the vertical component changes due to gravity.
The horizontal component of velocity (v_x) can be found using the following equation:
v_x = v * cos(θ)
where v is the initial velocity of the car.
The vertical component of velocity (v_y) can be found using the equation:
v_y = v * sin(θ)
At the highest point of the car's trajectory (when it leaves the cliff), the vertical component is zero since it has reached its maximum height. So, we can write:
v_y = 0
Using kinematic equations, we can determine the time it takes for the car to reach its maximum height. Let's consider the vertical motion only:
h = v_y * t + (1/2) * g * t^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since v_y = 0 at the highest point, the equation simplifies to:
h = (1/2) * g * t^2
Solving for t, we get:
t = sqrt((2 * h) / g)
Next, we can calculate the horizontal distance traveled by the car in time t using the equation:
x = v_x * t
Substituting the values we have, we get:
156. m = (v * cos(θ)) * sqrt((2 * 95. m) / 9.8 m/s^2)
Now, we can solve for v (the initial velocity of the car) by rearranging the equation:
v = 156. m / (cos(θ) * sqrt((2 * 95. m) / 9.8 m/s^2)
Plugging in the given values:
v ≈ 27.0 m/s
Therefore, the speed of the car as it left the cliff was closest to 27.0 m/s.